Let α and β be the roots of the quadratic equation x2 sin θ - x ( sin θ cos θ + 1) + cos θ = 0 ( 0

Question

Let α and β be the roots of the quadratic equation x2 sin θ - x ( sin θ cos θ + 1) + cos θ = 0 ( 0 < θ < 45° ), and α < β . Then displaystyle∑∞n = 0 ( αn + (( -1)n/βn)) is equal to : (A) (1/1 - cos θ) + (1/1+ sin θ)  (B) (1/1 + cos θ) + (1/1 + sin θ)  (C) (1/1 - cos θ) - (1/1+ sin θ)  (D) (1/1 + cos θ) - (1/1 - sin θ)

Tardigrade Admin · Accepted Answer

D =(1+ sinθ cosθ )2 - 4 sinθ cosθ =(1- sinθ cosθ )2 ⇒ roots are β = cos ec θ and α= cosθ ⇒ ∑∞n=0 (αn + (- (1/β))n ) = ∑∞n=0 ( cosθ)n + ∑nn=0 (- sinθ)n = (1/1- cosθ) + (1/1+ sinθ)

Q. Let $α$ and $β$ be the roots of the quadratic equation $x^{2} sin θ - x (sin θ cos θ + 1) + cos θ = 0 (0 < θ < 4 5^{\circ}),$ and $α < β$ . Then $n = 0 \sum \infty (α^{n} + \frac{( - 1 ) ^{n}}{β ^{n}})$ is equal to :

$\frac{1}{1 - c o s θ} + \frac{1}{1 + s i n θ}$

$\frac{1}{1 + c o s θ} + \frac{1}{1 + s i n θ}$

$\frac{1}{1 - c o s θ} - \frac{1}{1 + s i n θ}$

$\frac{1}{1 + c o s θ} - \frac{1}{1 - s i n θ}$

Q. Let α and β be the roots of the quadratic equation x2sinθ−x(sinθcosθ+1)+cosθ=0(0<θ<45∘), and α<β . Then n=0∑∞​(αn+βn(−1)n​) is equal to :

1−cosθ1​+1+sinθ1​

1+cosθ1​+1+sinθ1​

1−cosθ1​−1+sinθ1​

1+cosθ1​−1−sinθ1​

D=(1+sinθcosθ)2−4sinθcosθ=(1−sinθcosθ)2 ⇒ roots are β=cosecθ and α=cosθ ⇒∑n=0∞​(αn+(−β1​)n)=∑n=0∞​(cosθ)n+∑n=0n​(−sinθ)n =1−cosθ1​+1+sinθ1​

Q. Let $α$ and $β$ be the roots of the quadratic equation $x^{2} sin θ - x (sin θ cos θ + 1) + cos θ = 0 (0 < θ < 4 5^{\circ}),$ and $α < β$ . Then $n = 0 \sum \infty (α^{n} + \frac{( - 1 ) ^{n}}{β ^{n}})$ is equal to :

$\frac{1}{1 - c o s θ} + \frac{1}{1 + s i n θ}$

$\frac{1}{1 + c o s θ} + \frac{1}{1 + s i n θ}$

$\frac{1}{1 - c o s θ} - \frac{1}{1 + s i n θ}$

$\frac{1}{1 + c o s θ} - \frac{1}{1 - s i n θ}$